Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 7: Q45E (page 306)

Wire electrical-discharge machining (WEDM) is a process used to manufacture conductive hard metal components. It uses a continuously moving wire that serves as an electrode. Coating on the wire electrode allows for

cooling of the wire electrode core and provides an improved cutting performance. The article “HighPerformance Wire Electrodes for Wire ElectricalDischarge Machining—A Review” gave the following sample observations on total coating layer thickness (in mm) of eight wire electrodes used for

WEDM: \({\rm{21 16 29 35 42 24 24 25}}\)

Calculate a \({\rm{99\% }}\)CI for the standard deviation of the coating layer thickness distribution. Is this interval valid whatever the nature of the distribution? Explain.

Short Answer

Expert verified

The boundaries of the confidence interval for the standard deviation is \((4.8248,21.8461)\)

No, the population distribution is a normal distribution

Step by step solution

01

To Calculate a \({\rm{99\% }}\)CI for the standard deviation

Given:

\({\rm{c = 99\% = 0}}{\rm{.99}}\)

\(2116{\rm{ }}29{\rm{ }}35{\rm{ }}42{\rm{ }}24{\rm{ }}24{\rm{ }}25\)

The mean is the sum of all values divided by the number of values:

\({\rm{\bar x = }}\frac{{{\rm{21 + 16 + 29 + 35 + 42 + 24 + 24 + 25}}}}{{\rm{8}}}{\rm{ = }}\frac{{{\rm{216}}}}{{\rm{8}}}{\rm{ = 27}}\)

The variance is the sum of squared deviations from the mean divided by \({\rm{n - 1}}\).The standard deviation is the square root of the variance:

\({\rm{s = }}\sqrt {\frac{{{{{\rm{(21 - 27)}}}^{\rm{2}}}{\rm{ + \ldots }}{\rm{. + (25 - 27}}{{\rm{)}}^{\rm{2}}}}}{{{\rm{8 - 1}}}}} {\rm{\gg 8}}{\rm{.2115}}\)

02

To Determine the critical values

Determine the critical values using the chi-square table in the appendix, which are given in the row \({\rm{df = n - 1 = 8 - 1 = 7}}\)and in the columns of \(\frac{{{\rm{1 - c}}}}{{\rm{2}}}{\rm{ = 0}}{\rm{.005 and 1 - }}\frac{{{\rm{1 - c}}}}{{\rm{2}}}{\rm{ = 0}}{\rm{.995}}\) :

\(\begin{array}{l}{\rm{\chi }}_{{\rm{1 - 0}}{\rm{.005}}}^{\rm{2}}{\rm{ = \chi }}_{{\rm{0}}{\rm{.995}}}^{\rm{2}}{\rm{ = 0}}{\rm{.989 }}\\{\rm{\chi }}_{{\rm{0}}{\rm{.005}}}^{\rm{2}}{\rm{ = 20}}{\rm{.276}}\end{array}\)

The boundaries of the confidence interval for the standard deviation are then:

\(\begin{array}{l}\sqrt {\frac{{{\rm{n - 1}}}}{{{\rm{\chi }}_{{\rm{\alpha /2}}}^{\rm{2}}}}} {\rm{ \times s = }}\sqrt {\frac{{{\rm{8 - 1}}}}{{{\rm{20}}{\rm{.276}}}}} {\rm{ \times 8}}{\rm{.2115\gg 4}}{\rm{.8248}}\\\sqrt {\frac{{{\rm{n - 1}}}}{{{\rm{\chi }}_{{\rm{1 - \alpha /2}}}^{\rm{2}}}}} {\rm{ \times s = }}\sqrt {\frac{{{\rm{8 - 1}}}}{{{\rm{0}}{\rm{.989}}}}} {\rm{ \times 8}}{\rm{.2115\gg 21}}{\rm{.8461}}\end{array}\)

This interval is NOT valid whatever the nature of the distribution, because we require that the population distribution is a normal distribution.

Hence the boundaries of the confidence interval for the standard deviation is \((4.8248,21.8461)\)

No, the population distribution is a normal distribution

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Aphid infestation of fruit trees can be controlled either by spraying with pesticide or by inundation with ladybugs. In a particular area, four different groves of fruit trees are selected for experimentation. The first three groves are sprayed with pesticides 1, 2, and 3, respectively, and the fourth is treated with ladybugs, with the following results on yield:

Treatment ni = Number of Trees (Bushels/Tree) si

1 100 10.5 1.5

2 90 10.0 1.3

3 100 10.1 1.8

4 120 10.7 1.6

Let µi= the true average yield (bushels/tree) after receiving the ith treatment. Then

\({\bf{\theta = }}\frac{{\bf{1}}}{{\bf{3}}}{\bf{(}}{{\bf{\mu }}_{\bf{1}}}{\bf{ + }}{{\bf{\mu }}_{\bf{2}}}{\bf{ + }}{{\bf{\mu }}_{\bf{3}}}{\bf{) - }}{{\bf{\mu }}_{\bf{4}}}\)

measures the difference in true average yields between treatment with pesticides and treatment with ladybugs. When n1, n2, n3, and n4 are all large, the estimator \(\widehat {\bf{\theta }}\) obtained by replacing each \({{\bf{\mu }}_{\bf{i}}}\)by Xiis approximately normal. Use this to derive a large-sample 100(1 -α)% CI for \({\bf{\theta }}\), and compute the 95% interval for the given

data.

A journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural frequency (Hz) of delaminated beams of a certain type. The resulting interval was (229.764, 233.504). You decide that a confidence level of 99% is more appropriate than the 95% level used. What are the limits of the 99% interval? (Hint: Use the center of the interval and its width to determine \(\overline x \) and s.)

A normal probability plot of the n=\({\rm{26}}\) observations on escape time shows a substantial linear pattern; the sample mean and sample standard deviation are \({\rm{370}}{\rm{.69 and 24}}{\rm{.36}}\), respectively.

a. Calculate an upper confidence bound for population mean escape time using a confidence level of \({\rm{95\% }}\)

b. Calculate an upper prediction bound for the escape time of a single additional worker using a prediction level of \({\rm{95\% }}\). How does this bound compare with the confidence bound of part (a)?

c. Suppose that two additional workers will be chosen to participate in the simulated escape exercise. Denote their escape times by \({\rm{X27 and X28}}\), and let X new denote the average of these two values. Modify the formula for a PI for a single x value to obtain a PI for X new, and calculate a 95% two-sided interval based on the given escape data.

presented a sample of \({\rm{n = 153}}\)observations on ultimate tensile strength, the previous section gave summary quantities and requested a large-sample confidence interval. Because the sample size is large, no assumptions about the population distribution are required for the validity of the CI.

a. Is any assumption about the tensile-strength distribution required prior to calculating a lower prediction bound for the tensile strength of the next specimen selected using the method described in this section? Explain.

b. Use a statistical software package to investigate the plausibility of a normal population distribution.

c. Calculate a lower prediction bound with a prediction level of \({\rm{95\% }}\)for the ultimate tensile strength of the next specimen selected.

Determine the following:

a. The 95th percentile of the chi-squared distribution with \({\rm{v = 10}}\)

b. The 5th percentile of the chi-squared distribution with\({\rm{v = 10}}\)

\(\begin{array}{l}{\rm{c}}{\rm{.P}}\left( {{\rm{10}}{\rm{.98£ }}{{\rm{\chi }}^{\rm{2}}}{\rm{£ 36}}{\rm{.78}}} \right){\rm{,where }}{{\rm{\chi }}^{\rm{2}}}{\rm{ is achi - squared rv with \nu = 22}}\\{\rm{d}}{\rm{.P}}\left( {{{\rm{\chi }}^{\rm{2}}}{\rm{ < 14}}{\rm{.611}}} \right.{\rm{ or }}\left. {{{\rm{\chi }}^{\rm{2}}}{\rm{ > 37}}{\rm{.652}}} \right){\rm{,where }}{{\rm{\chi }}^{\rm{2}}}{\rm{ is achi squared rv with \nu = 25}}\end{array}\)
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free